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Can someone please explain to me how to prove this:

$\sin2\alpha = \frac{2\tan \alpha}{1+\tan^2 \alpha}$

Also: $\alpha \neq (2k + 1)\pi, k \in \mathbb{N}$

Thanks in advance.

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4 Answers 4

up vote 3 down vote accepted

$$\sin2\alpha=2\cos\alpha \sin\alpha=\dfrac{2\cos^2\alpha \sin\alpha}{\cos\alpha}=\dfrac{2\tan\alpha}{\sec^2\alpha}=\dfrac{2\tan\alpha}{1+\tan^2\alpha}$$

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Use the fact that $\sin{2a}=2\sin{a}\cos{a}$. Now what happens if you multiply by 1?

$\mathbf{hint}: 1=\frac{\cos{a}}{\cos{a}}$...

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Hint: Use the fact that $\sin 2\alpha = 2\sin \alpha \cos \alpha$ and $\sin^2 \alpha + \cos^2 \alpha = 1$. Use the last hint as the denominator of a fraction.

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Right hand side= 2 $Tanx/(1+tan^2x)$ substitute Tanx=$sinx/cosx$ and 1+$tan^2x$= $sec^2x$

2$(sinx/cosx)$/$sec^2x$= 2$(sinx/cosx)$ * $cos^2x$ = 2sinx* cosx + sin2x (left hand side )

proven

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